i printed and tested the jig, above. i mis-estimated the weight of the stock -- it doesn't need a string system to hold it, it can be held and rotated into position with the strength of a finger. excited to test its repeatability in the course of machining.

printing xmas presents (just kodi boxes)

what it's supposed to look like

what it looks like coming off the printer (wavy is b/c my x and y axes have poor repeatability... not built to exact constraint)

i sanded and then smoothed it using acetone. however, 3D printing is like making things with sticky rope, and between the ropes, there are crevices. this is where the acetone was hanging out when it was put in a rit bath. the acetone boiled out and created these bubbles of plastic along the crevices. it wasn't intentional but maybe it could be used to artistic effect. it messed up the dimensions unfortunately, but at least the hobby's cheap.

toyotathon posted:

i sanded and then smoothed it using acetone. however, 3D printing is like making things with sticky rope, and between the ropes, there are crevices. this is where the acetone was hanging out when it was put in a rit bath. the acetone boiled out and created these bubbles of plastic along the crevices. it wasn't intentional but maybe it could be used to artistic effect. it messed up the dimensions unfortunately, but at least the hobby's cheap.

that's cool

jansenist_drugstore posted:i've been reading about the replicability crisis in social sciences (only one problem of an obvious many). on the one hand, so-called experts are saying that this is not a problem of a blind faith in p-values, but other so-called experts are saying that p-values can no longer be reliably used in research. came across bayesian statistics, and am wondering if anyone has any thoughts/experience about this approach to research or its usefulness in general? is this just another attempt to rationalize poor methods?

during my final year of undergrad i took a biopsych module, and after attending lecture one to doublecheck there was no attendance register, i safely ignored the next 59. after a relaxing evening reading a few wikipedia articles on biological psychology i was pretty confident walking into my finals only imagine the momentary horror as i read on the front of the paper "failiure to reference real research will cap your maximum mark"; well, i thought, fortune favours the bold, so i just scattered a few (allen, allen & smith, 2002) and Johnson and Johnson (1987) through the paper; 64%; my greatest ever own

okay i looked up the formula, it is 180*(1 + 4f), where f is the fraction of the sphere covered by the triangle.

so if the triangle covers 1/4 it will = 360?, its not really a triangle then because one of the angles is a straight line at this point? like one side goes over the pole and another round the equator, so its just a big pacman mouth shape at this point, only got two sides...; i think ive jumped too far ahead in geometry with this, turns out i dont even know whata triangle is yet

the maximum case where it equals 900deg is interesting imo. it's where a triangle covers nearly the whole sphere. imagine one leg of the triangle growing from a point on an equator, looping all the way around, nearly touching the starting point. it is almost a full circle. to make this triangle have very little area (internal angle nearly 180deg) draw its other 2 legs at a very very shallow angle to the almost-circle. then connect them. i'm gonna call the pt where they connect 'A' cuz it will come up again. if they are shallow, their angles are like 0.001 degrees. the triangle area is very small, so internal angle is close to 180deg -- almost all the angle is from the last 2 legs you just drew, the angle at 'A', where they meet, cuz it's almost a flat line, 180 deg.

but now, take the two legs you just drew, disconnect them at 'A' where they meet, and spin each around, now have them meet at the backside -- all that area on the sphere that *wasn't* in the triangle, now *is* in the triangle. the triangle's gone from covering almost none of the sphere, to covering almost all of it. the "opposite" of a triangle on a sphere is also a triangle. according to the formula 180*(1+4f) a triangle that covers nearly all the sphere should have 900deg. 180*5=900. where did 900 come from tho. you have the angle where they meet, which is still 180. and you spun the 2 legs around 360deg each. 360+360+180=900 boom.

topologically you can make a 2D sphere out of a 2D triangle, by gluing edges in this way.

toyotathon posted:ain't math a kick

well heroditus says that pythagoras had some sort of proof for etenal life and stayed in his underground house for three years to the point where everyone thought he was dead then emerging in the fourth, so yes, i guess it has some charms

lenochodek posted:if you make the two triangles join a side in such a way they form a bigger triangle together you can have 3 w/o space remaining

oh i see, just divide the one in half; too simple, thanks.

tears posted:so then what happens when you draw a triangle on a 4d sphere?

this is a more complicated question since a triangle is really a 2d shape and a "4d" sphere (in the sense you mean) has 3 spatial degrees of freedom. you could still consider 2d triangles, in which case theres some ambiguity about how they would be embedded, or you could consider some related polyhedron, like a tetrahedron. im not sure about the exact answer to your question in those cases though, sorry....

when the isoceles triangle is very flat in the 5th image, area is nearly zero, and all the internal angles are around 0, 0, 180. as the point moves around the circle, along the equator, the area grows, and the two angles that were 0 grow to 180 each, when tri area is 1/2 sphere area, then when tri area is almost sphere area, the two angles which were once 0 have spun all the way around to 360. 360+360+180=900

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anyway maths is pretty funny but it leaves itself far too open for to me doing stupid things like: E = hf & E = mc2 so for the quantum of "earth", being the smallest possible number of earths (i.e. 1 earth, because you cant have <1 earth without having 0 earths, mass = 5.972 × 10²⁴ kg), f = 356138010.4 or whatever, i probably did the actual maths part wrong, but im still laughing

mΔths

toyotathon posted:

and this is cool, thanks, i understand

i think you are talking about a very useful tool in calculus called the 'limit', and not planck length, the ultramicronano speck of length is the 'infinitesimal'. a planck length's planck length divided by a trillion and then even a little bit smaller than that. marx wrote on the philosophy of the differential element some pgs back itt. it can help analyze functions -- like the investigation up there is a limit analysis of the min and max of the formula i found deg=180*(1+4f), that i didn't trust and needed to verify

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toyotathon posted:better explanation hopefully of the triangle internal angles on the sphere surface...

when the isoceles triangle is very flat in the 5th image, area is nearly zero, and all the internal angles are around 0, 0, 180. as the point moves around the circle, along the equator, the area grows, and the two angles that were 0 grow to 180 each, when tri area is 1/2 sphere area, then when tri area is almost sphere area, the two angles which were once 0 have spun all the way around to 360. 360+360+180=900

so try this one on then:

1. Take a sphere and inflate it to a huge degree

2. draw a tiny equilateral triangle on it, to the point that the curvature of the sphere is irelevant at the level of your triangle so all 3 angles of the tiny triangle equal 60, and so the angles of the triangle = 180

3. because you are on a sphere, you have automatically drawn a 2nd triangle encompasing the rest of the sphere, where each angle is 360-60 = 300, so you've drawn two triangles on a sphere with sum of angles 900 + 180 = 1080 degrees

i think this is basically the same thing, except its all triangles

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Edited by tears ()

tears posted:so then what happens when you draw a triangle on a 4d sphere?

a 2D triangle and a 2D sphere in 4D space will intersect at best at a single point, or if they are exactly equal, infinite points, but typically they will not intersect. i will try to demonstrate

in 1D (aka R1, the number line), looking at 2 points: (x) and (x') normally do not intersect, and if they do, it is at a single point, where x=x'. but normally x equals like, 5, and x' equals like 59871.333333. the typical case for 2 points in R1 is no intersection, and very rarely, intersection at a point, only when they're equal, x=x'.

in 1D, looking at 2 lines: two lines in 1D have to be equal, so there are infinitely many intersections. typical (and only) case is infinite intersections.

in 2D (aka R2, the xy plane), looking at 2 points: (x, 0) and (x', 0) is the same case as above, they only intersect when x=x'. and if the 2nd dimension isn't equal then there's no hope of two points ever intersecting. typical case for 2 points in R2 is no intersection, rare case is 1 intersection.

in 2D (R2), looking at 2 lines: if we up the dimension to look at lines on the plane, instead of points, it changes -- if you're reading euclid you know this. two lines normally intersect at 1 point. if the lines are parallel, they intersect at 0 points. if the lines are equal, they intersect at infinitely many points. but the typical case for 2 lines is 1 point of intersection, rare cases are 0 or infinite.

in 2D, looking at 2 planes: this is like 2 lines in 1D: there are typically (and only) infinitely many intersections.

in 3D (R3), looking at 2 points: (x, 0, 0) and (x', 0, 0), again, only are equal when x=x'. the typical case is no intersection, and at most, 1 intersection.

in 3D (R3), looking at 2 lines: this case is kind of like 2 zero-D points in R2, except it's 2 1D lines in R3. typical case is 2 lines in 3D never intersect. the rare cases: if they are in the same plane, it's the case above, 1 intersection, and if the lines are equal, infinitely many intersections.

in 3D, looking at 2 planes: this case is like 2D, looking at 2 lines. the typical case is that they intersect on a line. the rare cases: if the planes are parallel: 0 intersections, if equal, infinite intersections

in 3D, looking at 2 volumes: this case is like 2D looking at 2 planes, or 1D looking at 2 lines. two infinite volumes in the same space have typically (only) infinitely many intersections.

your question, what's the interaction between the 2D sphere and the 2D triangle in 4D:

2D is (x,y), 3D is (x,y,z), 4D we can use (x,y,z,w). so we can define a 2D sphere as (x,y,0,0) and a 2D triangle using (0,0,z,w). how many solutions are there for (x,y,0,0) = (0,0,z,w)? two 2D shapes in 4D will at best intersect at a single point. this is analogous to two 1D lines in 3D, or two 0D points in 2D. if the 2D sphere and 2D triangle are (nearly, in the limits i drew in the pic) equal then infinitely many points.

don't ask me anymore questions about this b/c this is all i know lol but maybe another person can help

toyotathon posted:your question, what's the interaction between the 2D sphere and the 2D triangle in 4D:

2D is (x,y), 3D is (x,y,z), 4D we can use (x,y,z,w). so we can define a 2D sphere as (x,y,0,0) and a 2D triangle using (0,0,z,w). how many solutions are there for (x,y,0,0) = (0,0,z,w)? two 2D shapes in 4D will at best intersect at a single point. this is analogous to two 1D lines in 3D, or two 0D points in 2D. if the 2D sphere and 2D triangle are (nearly, in the limits i drew in the pic) equal then infinitely many points.

don't ask me anymore questions about this b/c this is all i know lol but maybe another person can help

The mistake you're making is that you're defining what you call a "2d sphere" and a "2d triangle" explicitly so that the only possibility is for them to intersect at the origin. Here's a counterexample, with a "2d sphere" intersecting a "2d triangle" everywhere on the triangle:

2d sphere: all points of the form (a,b,0,0) where the distance of the point from origin is less than 2.

2d triangle: all points in the region bounded by lines between (0,0,0,0), (1,0,0,0), and (0,1,0,0)

We can agree that the shapes are right. Clearly every point in the triangle lies within the sphere.

Practically speaking, if you look at the area where these two shapes intersect in 3d, that basically describes the kinds of areas they can intersect in any larger dimension. They both sit on planes by definition. When you add a third dimension to move them around in, they don't have to be coplanar anymore and you can have them intersect in a line if you want, which you can't do without that third dimension. After 3, the other dimensions don't really add anything. The options are: they intersect nowhere, they intersect at exactly one point (when the edge of a triangle touches the edge of the circle), they intersect along a line (in 3d and above), and they intersect on a plane in whatever weird shapes you can think of.

(x,y,0,0) = (0,0,z,w) where the 2D triangle's defined by x,y and the 2D sphere by z,w.. this works also if you put the triangle in another plane (x,8,z,0) in R4 and the sphere (0,8,z',w), there is still only one solution, 1 pt of intersection, x=0, z=z', w=0. if it's (x,8,z,0) and (0,7,z',w) then there is no solution where they are equal, no intersection, since 8 != 7

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I think in common language we tend to use the the words sphere and ball interchangeably.

It might be helpful to forget the analytic approach and visualize. To imagine a 2-sphere and a triangle intersecting in R4, imagine them in R3 first, but with two different colors. Color will stand in for our fourth spatial coordinate. Lets say the triangle is yellow and the sphere is red. Imagine them touching. If the intersections don't share the same color, they are not the same point in R4: yellow does not equal red, so the fourth coordinate is not identical.

Now imagine the yellow triangle remaining in place but color shifting from yellow through orange to red. Intersect the two shapes however you wish. All configurations where they intersect in this 'both red' space are true intersections in R4.

It seems clear to me that any intersections we can achieve in R3 can also be achieved in R4, if we keep the fourth coordinate, color, identical. Thus, they can intersect along at a point, a line, or not at all

i don't feel comfortable w/ my 4D intuition so i have to do algebra... back down in R2, two lines intersect when their equations are equal. so like line 1: f(x) = 4*x + 8, line 2: g(x') = x' - 4. these lines are defined by (x, f(x)) and (x', g(x')), and they are equal when (x,f(x)) = (x',g(x')) = (a, b). they are 1D objects since there is only 1 free parameter, x or x'. the solution of their intersection can be found when f(x) = g(x') = b, and x=x'=a, and (a,b) is the pt of intersection. 4*x+8 = x-5, x = x' = a = -3/12 = -1/4, f(x)=g(x)=b=-1+8=7, so we can solve the intersection (just one point) at (a,b)=(-1/4, 7). that's the dimension of the solution space... 1D + 1D in R2 is 0D, and rarely, 1D (f(x) and g(x) are the same equation)) or no solution (slopes are equal). but if you were to pick slopes and y-intercepts at random from the reals, the odds of landing on either of the two 'rare' cases (no solution and 1D) are exactly zero.

if the 2-sphere is in R4, but not really 'using' the 4th dim, w/ spherical coords like (5, lat, long, 420) and a 2D disk with (rad, 0, long, 420), it is the R3 case, intersection along the 2-sphere's equator. (5, lat, long, 420)=(rad, 0, long, 420) is true when rad=5, lat=0, long is the 1D free parameter, w=420, so the solution space is 1D. not 'using' w and moving the disk into another plane, (rad, lat, 0, 420), the solution space is the prime meridian. the most typical solution tho is a point. if the disk is in (rad, 0, 420, long), it can be verified that the solution is a point. if the sphere is a ball, defined by (rad, lat, long, 420), the solution space will be rarely a plane and typically a line...

this is fun stuff cuz the dimensionality of the solution space is directly related to Exact Constraint theory on pg1, the intersection of kinematic spaces in R3+R3. the dimension of that intersection is how many DOFs there are.

toyotathon posted:don't ask me anymore questions about this b/c this is all i know lol but maybe another person can help

thats ok, i'll watchcube 2: hupercube again, followed by its sequel cube zero

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Edited by toyotathon ()

Belphegor posted:That would be a good notation for replacing degrees. Degrees are essentially a remnant of Babylonian sexagesimal notation when pretty much everything else is decimal. Were stuck with them. The properties of your notation would be very helpful for trigonometry education!

We already have radians for more "theoretical" applications, but the advantages of having base 360 is that the common angles are all integer values: e.g. 15, 30, 45, 60, etc, because 360 = 2^3 * 3^2 * 5 (notice that these are descending powers of the first 3 prime numbers and 100 = 2^2*5^2 missing the power of 3), whereas if you had 1 deg be = 2pi radians, you would have to write all the common angles such as 2pi/3 = 0.3333... deg, etc. Furthermore, you wouldn't have the easy conversions that radians give you either e.g. the arclength of a circle would no longer just be theta*radius. Obviously, it's just a linear transformation, so it wouldn't be a huge deal either way, but there's little you're gaining with the conversion. Anecdotally, teaching radians to kids is harder than angles. There seems to be a bit of a conceptual hurdle to mapping the interval [0,2pi) to an angle, and I'm not sure [0,1) would improve that much. Obviously degrees are the same thing, however, it's relatively easy to show a picture of a circle split into 360 slices to illustrate how degrees work.

Radians were way harder to learn than degrees, I probably learned degrees 10 years before radians.

elemennop posted:Belphegor posted:

We already have radians for more "theoretical" applications, but the advantages of having base 360 is that the common angles are all integer values: e.g. 15, 30, 45, 60, etc, because 360 = 2^3 * 3^2 * 5 (notice that these are descending powers of the first 3 prime numbers and 100 = 2^2*5^2 missing the power of 3), whereas if you had 1 deg be = 2pi radians, you would have to write all the common angles such as 2pi/3 = 0.3333... deg, etc. Furthermore, you wouldn't have the easy conversions that radians give you either e.g. the arclength of a circle would no longer just be theta*radius. Obviously, it's just a linear transformation, so it wouldn't be a huge deal either way, but there's little you're gaining with the conversion. Anecdotally, teaching radians to kids is harder than angles. There seems to be a bit of a conceptual hurdle to mapping the interval [0,2pi) to an angle, and I'm not sure [0,1) would improve that much. Obviously degrees are the same thing, however, it's relatively easy to show a picture of a circle split into 360 slices to illustrate how degrees work.

ya i prefer to work in degrees cuz of all the factors up to 15 (didn't realize its divisibility came out of its prime factors that's cool, makes sense) but check it out: what's more obvious to a kid, 270°, 3pi/2, or 3/4° = 3/4 of a circle?? i think that, kids could learn modular arithmetic w/ the clock really fast. like 13/12° --> subtract the bottom from the top --> (13-12)/12 ---> 1/12 of the circle = 1 o'clock. keep subtracting until top < bottom. it is also arguably more "metric"... angle rn is a*360 degree + b/60 minutes + c/3600 seconds, mod 360,60,60... silly units... but a fraction of 1° could be naturally either like 24/25° or a decimal 0.928121212°, base 10 and mod 1. reducing an angle like 747° to 27°, ppl don't gotta remember factors of 360, lop off the left-hand side of the decimal: 2.978° = 0.978°. and if you wanted the arc length it's 2pi*0.978... that's the conversion to radians.. and the same divisibility of 360 helps the other way, 1/3 big degree * 360 old degrees = 120 old degrees.

edit: for negative angles expressed as fractions, mod1 has another nice property, if you take say, -5/27°, put the bottom on the top: -5/27 = (27-5)/27 = 22/27°. this is because x/n + n/n + n/n + ... = x/n in mod 1, so -5/27 + 27/27 = (27-5)/27. very quick calc vs 360 or radians! and to go back to the sine/cosine functions, rn the "box" that surrounds one period of a*sin(c*x), for degrees is 2a : 360c (2 amplitudes and the wavelength), for radians is 2*a : 2pi*c, but for the 1° circle it is a half-square, 2:1... the y range = 2*the x range in one period, for drawing the function on graph paper w/ easy scaling of x and y. fwiw i am eating where i shit and am calculating using this, b/c it is a natural way to parametrize a sheet on the real plane (s,t) s in (0,1) and t in (0,1) where one side of the plane reaches around and attaches to the other side as a cylinder, where s(1) = s(0) or t(1) = t(0), stitching it like a blanket's opposite ends. anyway i think it's a better system, and even if it weren't, it's my amerikkkan right to stubbornly use ridiculous units!!!!!!

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